t-test practice problems (one-sample and paired)

State. Let $\mu$ be the machine's true mean fill in milliliters. $H_0: \mu = 500$ versus $H_a: \mu \ne 500$, with $\alpha = 0.05$.

Plan. One-sample t-test for a mean. Random: the 16 bottles are a random sample. 10%: 16 is less than 10% of all bottles the machine fills. Normal/large sample: with $n = 16 < 30$, assume the fill amounts are roughly normal with no strong skew or outliers.

Find the standard error. $SE = \frac{s}{\sqrt{n}} = \frac{8}{\sqrt{16}} = \frac{8}{4} = 2$ mL.

Find the test statistic and degrees of freedom. $t = \frac{\bar{x} - \mu_0}{SE} = \frac{497 - 500}{2} = \frac{-3}{2} = -1.50$, with $df = n - 1 = 15$.

Bound the p-value. This is two-sided, so use $|t| = 1.50$. In the $df = 15$ row, $1.50$ falls between $1.341$ (tail $0.10$) and $1.753$ (tail $0.05$), so one tail holds between $0.05$ and $0.10$. Doubling, the two-sided p-value is between $0.10$ and $0.20$.

Conclude. Since the p-value is greater than $0.10$, which exceeds $0.05$, fail to reject $H_0$. There is not convincing evidence that the machine's true mean fill differs from 500 mL.

State. Let $\mu$ be the true mean cook time in minutes. $H_0: \mu = 30$ versus $H_a: \mu > 30$, with $\alpha = 0.05$.

Plan. One-sample t-test. Random: the 25 recipes are a random sample. 10%: 25 is less than 10% of all the company's recipes. Normal/large sample: with $n = 25 < 30$, assume the cook times are roughly normal with no strong skew or outliers.

Find the standard error. $SE = \frac{s}{\sqrt{n}} = \frac{7.5}{\sqrt{25}} = \frac{7.5}{5} = 1.5$ minutes.

Find the test statistic and degrees of freedom. $t = \frac{33 - 30}{1.5} = \frac{3}{1.5} = 2.00$, with $df = 25 - 1 = 24$.

Bound the p-value. Right-tailed. In the $df = 24$ row, $2.00$ falls between $1.711$ (tail $0.05$) and $2.064$ (tail $0.025$), so the p-value is between $0.025$ and $0.05$.

Conclude. Since the p-value is less than $0.05$, reject $H_0$. There is convincing evidence that the true mean cook time exceeds 30 minutes.

State. Let $\mu$ be the kiln's true mean peak temperature in degrees Fahrenheit. $H_0: \mu = 1200$ versus $H_a: \mu \ne 1200$, with $\alpha = 0.05$.

Plan. One-sample t-test. Random: the 9 firings are a random sample. 10%: 9 is less than 10% of all firings. Normal/large sample: with $n = 9 < 30$, assume the peak temperatures are roughly normal with no strong skew or outliers.

Find the standard error. $SE = \frac{s}{\sqrt{n}} = \frac{15}{\sqrt{9}} = \frac{15}{3} = 5$ degrees.

Find the test statistic and degrees of freedom. $t = \frac{1215 - 1200}{5} = \frac{15}{5} = 3.00$, with $df = 9 - 1 = 8$.

Bound the p-value. Two-sided, so use $|t| = 3.00$. In the $df = 8$ row, $3.00$ falls between $2.896$ (tail $0.01$) and $3.355$ (tail $0.005$), so one tail holds between $0.005$ and $0.01$. Doubling, the two-sided p-value is between $0.01$ and $0.02$.

Conclude. Since the p-value is less than $0.02$, which is below $0.05$, reject $H_0$. There is convincing evidence that the kiln's true mean peak temperature differs from 1200°F.

State. Let $\mu_d$ be the true mean increase in grip strength (after minus before). $H_0: \mu_d = 0$ versus $H_a: \mu_d > 0$, with $\alpha = 0.05$.

Plan. Paired t-test on the 7 differences. Random: the 7 patients are a random sample. 10%: 7 is less than 10% of all such patients. Normal/large sample: with $n = 7 < 30$, assume the differences are roughly normal with no strong skew or outliers.

Find the mean difference. $\bar{d} = \frac{1 + 3 + 0 + 2 + 4 + 1 + 3}{7} = \frac{14}{7} = 2$ kg.

Find the standard deviation of the differences. The squared deviations from 2 are $1, 1, 4, 0, 4, 1, 1$, summing to $12$, so $s_d = \sqrt{\frac{12}{7 - 1}} = \sqrt{\frac{12}{6}} = \sqrt{2} = 1.414$ kg.

Find the standard error, test statistic, and df. $SE = \frac{s_d}{\sqrt{n}} = \frac{1.414}{\sqrt{7}} = \frac{1.414}{2.646} = 0.535$, so $t = \frac{\bar{d}}{SE} = \frac{2}{0.535} = 3.74$ with $df = 7 - 1 = 6$.

Bound the p-value. Right-tailed. In the $df = 6$ row, $3.74$ falls between $3.707$ (tail $0.005$) and $4.317$ (tail $0.0025$), so the p-value is between $0.0025$ and $0.005$.

Conclude. Since the p-value is less than $0.005$, which is below $0.05$, reject $H_0$. There is convincing evidence that the program increases mean grip strength.

State. Let $\mu_d$ be the true mean difference in readings (Brand A minus Brand B). $H_0: \mu_d = 0$ versus $H_a: \mu_d \ne 0$, with $\alpha = 0.05$.

Plan. Paired t-test on the 6 differences, one per volunteer measured on both monitors. Random: the problem does not state the 6 volunteers were randomly selected, so treat them as a volunteer (convenience) sample; the conclusion applies to these volunteers and does not generalize to a wider population. Normal/large sample: with $n = 6 < 30$, assume the differences are roughly normal with no strong skew or outliers.

Find the mean difference. $\bar{d} = \frac{2 + (-1) + 3 + (-2) + 4 + 0}{6} = \frac{6}{6} = 1$ mmHg.

Find the standard deviation of the differences. The squared deviations from 1 are $1, 4, 4, 9, 9, 1$, summing to $28$, so $s_d = \sqrt{\frac{28}{6 - 1}} = \sqrt{5.6} = 2.366$ mmHg.

Find the standard error, test statistic, and df. $SE = \frac{s_d}{\sqrt{n}} = \frac{2.366}{\sqrt{6}} = \frac{2.366}{2.449} = 0.966$, so $t = \frac{\bar{d}}{SE} = \frac{1}{0.966} = 1.04$ with $df = 6 - 1 = 5$.

Bound the p-value. Two-sided, so use $|t| = 1.04$. In the $df = 5$ row, $1.04$ falls between $0.920$ (tail $0.20$) and $1.156$ (tail $0.15$), so one tail holds between $0.15$ and $0.20$. Doubling, the two-sided p-value is between $0.30$ and $0.40$.

Conclude. Since the p-value is greater than $0.30$, which exceeds $0.05$, fail to reject $H_0$. There is not convincing evidence that the two monitor brands give different mean readings.

(a) Let $\mu$ be the true mean weight of the packs in grams. Because the group suspects underfilling, $H_0: \mu = 25$ versus $H_a: \mu < 25$, with $\alpha = 0.05$.

(b) One-sample t-test. Random: the 12 packs are a random sample. 10%: 12 is less than 10% of all packs produced. Normal/large sample: with $n = 12 < 30$, assume the weights are roughly normal with no strong skew or outliers.

(c) Standard error. $SE = \frac{s}{\sqrt{n}} = \frac{1.2}{\sqrt{12}} = \frac{1.2}{3.464} = 0.346$ g.

(c) Test statistic and degrees of freedom. $t = \frac{\bar{x} - \mu_0}{SE} = \frac{24.2 - 25}{0.346} = \frac{-0.8}{0.346} = -2.31$, with $df = 12 - 1 = 11$.

(c) p-value. Left-tailed, so use $|t| = 2.31$. In the $df = 11$ row, $2.31$ falls between $2.201$ (tail $0.025$) and $2.328$ (tail $0.02$), so the p-value is between $0.02$ and $0.025$.

(d) Conclude. Since the p-value is less than $0.025$, which is below $0.05$, reject $H_0$. There is convincing evidence that the packs' true mean weight is less than 25 g, supporting the underfilling claim.

(a) Let $\mu_d$ be the true mean yield difference (treated minus untreated). $H_0: \mu_d = 0$ versus $H_a: \mu_d \ne 0$, with $\alpha = 0.05$. This is a matched-pairs design: plots are paired by location and the additive is randomly assigned within each pair, so inference rests on that random assignment. With $n = 8 < 30$, assume the differences are roughly normal with no strong skew or outliers.

(b) Mean difference. $\bar{d} = \frac{2 + 0 + 3 + (-1) + 1 + 4 + 1 + 2}{8} = \frac{12}{8} = 1.5$ kg.

(b) Standard deviation of the differences. The squared deviations from 1.5 are $0.25, 2.25, 2.25, 6.25, 0.25, 6.25, 0.25, 0.25$, summing to $18$, so $s_d = \sqrt{\frac{18}{8 - 1}} = \sqrt{2.571} = 1.604$ kg.

(b) Standard error, test statistic, and df. $SE = \frac{s_d}{\sqrt{n}} = \frac{1.604}{\sqrt{8}} = \frac{1.604}{2.828} = 0.567$, so $t = \frac{\bar{d}}{SE} = \frac{1.5}{0.567} = 2.65$ with $df = 8 - 1 = 7$.

(b) p-value. Two-sided, so use $|t| = 2.65$. In the $df = 7$ row, $2.65$ falls between $2.517$ (tail $0.02$) and $2.998$ (tail $0.01$), so one tail holds between $0.01$ and $0.02$. Doubling, the two-sided p-value is between $0.02$ and $0.04$.

(c) Conclude at $\alpha = 0.05$. Since the p-value is less than $0.04$, which is below $0.05$, reject $H_0$. There is convincing evidence that the additive changes mean yield.

(d) At $\alpha = 0.01$, the p-value is between $0.02$ and $0.04$, which is greater than $0.01$, so you fail to reject $H_0$. The conclusion changes: at the stricter level there is not convincing evidence of a change in mean yield.

(a) Let $\mu_d$ be the true mean change in reaction time (restricted minus normal); because researchers expect restricted sleep to slow reactions, $H_0: \mu_d = 0$ versus $H_a: \mu_d > 0$ at $\alpha = 0.05$. Paired t-test on the 15 differences, each participant measured under both conditions. Random: the 15 participants are a random sample; 10%: 15 is less than 10% of all people the sample represents. Normal/large sample: with $n = 15 < 30$, assume the differences are roughly normal with no strong skew or outliers.

(b) Standard error. $SE = \frac{s_d}{\sqrt{n}} = \frac{20}{\sqrt{15}} = \frac{20}{3.873} = 5.164$ ms.

(b) Test statistic and degrees of freedom. $t = \frac{\bar{d}}{SE} = \frac{12}{5.164} = 2.32$, with $df = 15 - 1 = 14$.

(b) p-value. Right-tailed. In the $df = 14$ row, $2.32$ falls between $2.264$ (tail $0.02$) and $2.624$ (tail $0.01$), so the p-value is between $0.01$ and $0.02$.

(c) Conclude. Since the p-value is less than $0.02$, which is below $0.05$, reject $H_0$. There is convincing evidence that restricted sleep increases mean reaction time.

(d) Interpret. If restricted sleep truly had no effect on mean reaction time, there would be between a 1% and 2% chance of getting a sample mean difference at least as large as 12 ms in the slower direction from random variation alone. Because you rejected $H_0$, the decision could be a Type I error: concluding restricted sleep slows reactions when in truth it makes no difference.